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Xelga [282]
3 years ago
7

A sock drawer has 8 blue socks and 6 black socks. If I randomly select two socks, one at a time, what is the probability I will

first get a blue sock and then, without replacing it, a black sock?
A.) 24/91

B.) 2/7

C.) 6/81

D.) 4/13
Mathematics
1 answer:
MAVERICK [17]3 years ago
6 0

The probability is most likely going to be

4/13

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The sum of three consecutive numbers is 276. What is the smallest of these intengers?
shtirl [24]

Answer:

91

Step-by-step explanation:

Let x be the smallest one:

● x is the first number

● x+1 is the second number

● x+2 is the third number

The sum of these numbers is 276

● x+(x+1)+(x+2) =276

● x+x+1+x+2 = 276

● 3x + 3 = 276

Substract 3 from both sides:

● 3x+3-3 = 276-3

● 3x = 273

Divide both sides by 3

● (3x)/3 = 273/3

● x = 91

So the smallest one is 91

5 0
3 years ago
What is the ratio of the number of days in the week that begin with the letter "T" to the number of days in the week that do not
ioda

Answer:

2:5

Step-by-step explanation:

3 0
2 years ago
Read 2 more answers
How many dots will figure n have? Explain how you know this.
LenKa [72]
N = ((n-1) + 4) + 4, to my understanding. 
although your answers seem right written in the columns, but your method of applying numbers to equation seems incorrect in 'expand' column. for example, answer for 3 should be something like this...
3   21   17+4
4   25    21+4
hope it helps.
5 0
3 years ago
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A laundry basket contains 18 blue socks and 24 black socks. What is the probability of randomly picking 2 black socks, without r
lisov135 [29]
P(black socks): 24/42 or 12/21
P(black socks without replacing): 23/41. As a result, the probability of randomly picking 2 black socks, without replacement, from the basket is 12/21×23/41=276/861 or 32%. Hope it help!
6 0
3 years ago
Reading proficiency: An educator wants to construct a 99.5% confidence interval for the proportion of elementary school children
Shkiper50 [21]

Answer:

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of \pi, and a confidence level of 1-\alpha, we have the following confidence interval of proportions.

\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}

In which

z is the zscore that has a pvalue of 1 - \frac{\alpha}{2}.

The margin of error of the interval is given by:

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

In this problem, we have that:

p = 0.69

99.5% confidence level

So \alpha = 0.005, z is the value of Z that has a pvalue of 1 - \frac{0.005}{2} = 0.9975, so Z = 2.81.

Using this estimate, what sample size is needed so that the confidence interval will have a margin of error of 0.07?

This is n when M = 0.07. So

M = z\sqrt{\frac{\pi(1-\pi)}{n}}

0.07 = 2.81\sqrt{\frac{0.69*0.31}{n}}

0.07\sqrt{n} = 1.2996

\sqrt{n} = \frac{1.2996}{0.07}

\sqrt{n} = 18.5658

(\sqrt{n})^{2} = (18.5658)^{2}

n = 345

A sample size of 345 is needed so that the confidence interval will have a margin of error of 0.07

4 0
3 years ago
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